Sampling Distribution
Yuanting Lu
11/10/2025
Sampling Distribution
“One of the key concerns of statistics is the drawing of conclusions from a set of observed data. These data will usually consist of a sample of certain elements of a population, and the objective will be to use the sample to draw conclusions about the entire population.”
In this lecture, we will learn about how to construct distributions of sample statistics (e.g., minimum, maximum, mean, median, proportion, standard deviation).
- Determine which parameter we would like to know from the population.
- Draw a random sample of size \(n\).
- Calculate the sample statistic.
- Repeat steps 2 and 3 a large number of times.
- Display all the sample statistic we obtained in step 4 on the same graph.
Central Limit Theorem
Central Limit Theorem:
For a random sample of size \(n\) from a population with mean \(\mu\) and standard deviation \(\sigma\), the sampling distribution of the sample mean, \(\overline{X}\), is approximately normal and has a mean of \(\mu\) and a standard deviation of \(\dfrac{\sigma}{\sqrt n}\).
\[E(\overline{X})=\mu, \mbox{ and }SD(\overline{X}\,)=\dfrac{\sigma}{\sqrt n}.\]
“… practically speaking, no matter how nonnormal the underlying population distribution is, the sample mean of a sample size of at least 30 will be approximately normal.”
Suppose \(X\) and \(Y\) are independent normal random variables. The additive properties are
\[E[X + Y] = E[X] + E[Y],\] \[Var[X + Y] = Var[X] + Var[Y].\]
The constant multiple properties are \[E[\color{red}cX] = \color{red}cE[X],\] \[Var[\color{red}cX] = \color{red}{c^2} Var[X].\]
If the independent random variables \(X_1, X_2, \ldots, X_n\) are from the same population, whose mean is \(\mu\) and standard deviation is \(\sigma\), then
- \(E[X_1]=E[X_2]=\cdots=E[X_n]=\mu\).
- \(Var[X_1]=Var[X_2]=\cdots=Var[X_n]=\sigma^2\).
Therefore, \[E[\overline{X}] = E\left[\dfrac{X_1 + X_2 + \cdots + X_n}{n}\right] = \dfrac{E[X_1]+E[X_2] + \cdots + E[X_n]}{n}=\dfrac{n\mu}{n}=\mu.\]
\[Var[\overline{X}] = Var\left[\dfrac{X_1 + X_2 + \cdots + X_n}{n}\right] = \dfrac{Var[X_1]+Var[X_2] + \cdots + Var[X_n]}{n^2}=\dfrac{n\sigma^2}{n^2}=\dfrac{\sigma^2}{n}.\]
\[SD(\overline{X})=\sqrt{Var(\overline{X})}=\dfrac{\sigma}{\sqrt{n}}.\]
If 1 man is randomly selected, find the probability that his weight is less than 167 lb.
If 36 men are randomly selected, find the probability that their average weight is less than 167 lb.
If 1 man is randomly selected, find the probability that his weight is between 170 and 175 lb.
If 64 men are randomly selected, find the probability that their mean weight is between 170 and 175.
You are to design an elevator to safely hold 16 people. Find the maximum allowable weight if we want a 0.95 probability that this maximum will not be exceeded in the worst case when 16 randomly selected males are on it.
Sampling Distribution of Sample Proportion
Suppose that the underlying population is large in relation to the sample size \(n\). If the proportion of individuals in the population with a certain characteristic is \(p\), then the sampling distribution of sample proportion (\(\widehat{p}\)) is approximately normal, \[E[\widehat{p}] = p, \quad\mbox{ and }\quad SD(\widehat{p}) = \sqrt{\dfrac{p(1-p)}{n}}.\]
A rule of thumb is that when np(1-p)>10, the binomial distribution can be approximated by a normal distribution.
When the underlying population size is way larger than the sample size, we can assume that the probability of each individual in the sample with the characteristic is \(p\). If we denote
- 1 for with the characteristic;
- 0 for the lack of the characteristic,
then the distribution for each individual is
| \(X_i\) | \(P(X_i)\) |
| 1 | \(p\) |
| 0 | \(1-p\) |
The expected value is \(E[X_i]=(1)(p) + (0)(1-p)=p\) and it follows that the variance is \[Var(X_i) = (1-E[X_i])^2(p) + (0 - E[X_i])^2(1-p)=(1-p)^2p+(0-p)^2(1-p)=p(1-p).\]
Therefore, \[E[\widehat{p}]=E\left[\dfrac{X_1 + X_2 + \cdots + X_n}{n}\right] = \dfrac{E[X_1]+E[X_2]+\cdots+E[X_n]}{n}=\dfrac{np}{n}=p.\]
\[\small Var(\widehat{p})=Var\left(\dfrac{X_1 + X_2 + \cdots + X_n}{n}\right)=\dfrac{Var(X_1)+Var(X_2)+\cdots+Var(X_n)}{n^2}=\dfrac{np(1-p)}{n^2}=\dfrac{p(1-p)}{n}.\]
\[SD(\widehat{p})=\sqrt{Var(\widehat{p})}=\sqrt{\dfrac{p(1-p)}{n}}.\]
The data file ATL_Departure_Flights_2017.csv has the
flights status information (on-time or delayed) of all the domestic
departure flights in Atlanta Hartsfield-Jackson Airport 2017.
How large is the dataset?
What percentage of departure flights were on-time in 2017?
Take a random sample of 50 flights. What percentage of departure flights were on-time in the sample?
Repeatedly take 30 or more samples and create a distribution graph.
To draw 2000 samples and create a simulated distribution graph.
n <- 50
pile <- rep(0, 2000)
for (i in 1:length(pile)) {
x <- departure$Status[sample(364655, n)]
phat <- table(x) / n
pile[i] <- as.numeric(phat[2])
}
stripchart(pile, method = 'stack', pch = 19,
at = 0.15, offset = 0.02, xlim = c(0, 1),
main = "Sampling Distribution of Sample Proportion (n=50)",
xlab = "Proportion of on-time departure flights")
- What is the probability that the proportion of the on-time flights is larger than 0.9?
- What is the probability that less than half of the flights depart on time?
- What is the probability that the proportion of the on-time flights is between 0.8 and 0.85?
Continuous Correction
The possible proportions in a particular sample are
But, the normal distribution is continuous. Therefore, the following
rules are called the
| Binomial Probability | Normal Approximation |
|---|---|
| \(P(X=a)\) | \(P(a - 0.5 < X < a + 0.5)\) |
| \(P(X>a)\) or \(P(X\ge a + 1)\) | \(P(X > a + 0.5)\) |
| \(P(X<a)\) or \(P(X\le a - 1)\) | \(P(X < a - 0.5)\) |
